How do you divide #(4+2i )/( 1-i)#?

2 Answers
Feb 8, 2016

Answer:

# 1+3i#

Explanation:

You must eliminate the complex number in the denominator by multiplying by its conjugate:

#(4+2i)/(1-i)=((4+2i)(1+i))/((1-i)(1+i))#

# (4+4i +2i+2i^2)/(1-i^2)#

#(4+6i-2)/(1+1)#

#(2+6i)/2#

# 1+3i#

Feb 8, 2016

Answer:

1 + 3i

Explanation:

Require the denominator to be real . To achieve this multiply the numerator and the denominator by the complex conjugate of the denominator.

If (a + bi ) is a complex number then (a - bi) is the conjugate

here the conjugate of (1 - i ) is (1 + i )

now # ((4 + 2i )(1 + i ))/((1 - i)(1 + i )) #

distribute the brackets to obtain :

# (4 + 6i + 2i^2 )/(1 - i^2 ) #

note that # i^2 = (sqrt(-1)^2 )= - 1 #

hence # (4 + 6i - 2 )/(1 + 1 ) = (2 + 6i )/2 = 2/2 + (6i)/2 = 1 + 3i#